Whakaoti i te 10y^2+19y+6

$$10 { y }^{ 2 } +19y+6$$

Tauwehe

$$\left(2y+3\right)\left(5y+2\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 10y^{2}+ay+by+6. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=19 ab=10\times 6=60$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōrunga te a+b, he tōrunga hoki a a me b. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 60.

$$1,60 2,30 3,20 4,15 5,12 6,10$$

Tātaihia te tapeke mō ia takirua.

$$1+60=61 2+30=32 3+20=23 4+15=19 5+12=17 6+10=16$$

Ko te otinga te takirua ka hoatu i te tapeke 19.

$$a=4 b=15$$

Tuhia anō te 10y^{2}+19y+6 hei \left(10y^{2}+4y\right)+\left(15y+6\right).

$$\left(10y^{2}+4y\right)+\left(15y+6\right)$$

Tauwehea te 2y i te tuatahi me te 3 i te rōpū tuarua.

$$2y\left(5y+2\right)+3\left(5y+2\right)$$

Whakatauwehea atu te kīanga pātahi 5y+2 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(5y+2\right)\left(2y+3\right)$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Ka taea te huamaha pūrua te tauwehe mā te whakamahi i te huringa ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), ina ko x_{1} me x_{2} ngā otinga o te whārite pūrua ax^{2}+bx+c=0.

$$10y^{2}+19y+6=0$$

Ko ngā whārite katoa o te āhua ax^{2}+bx+c=0 ka taea te whakaoti mā te whakamahi i te tikanga tātai pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. E rua ngā otinga ka puta i te tikanga tātai pūrua, ko tētahi ina he tāpiri a ±, ā, ko tētahi ina he tango.

$$y=\frac{-19±\sqrt{19^{2}-4\times 10\times 6}}{2\times 10}$$

Pūrua 19.

$$y=\frac{-19±\sqrt{361-4\times 10\times 6}}{2\times 10}$$

Whakareatia -4 ki te 10.

$$y=\frac{-19±\sqrt{361-40\times 6}}{2\times 10}$$

Whakareatia -40 ki te 6.

$$y=\frac{-19±\sqrt{361-240}}{2\times 10}$$

Tāpiri 361 ki te -240.

$$y=\frac{-19±\sqrt{121}}{2\times 10}$$

Tuhia te pūtakerua o te 121.

$$y=\frac{-19±11}{2\times 10}$$

Whakareatia 2 ki te 10.

$$y=\frac{-19±11}{20}$$

Nā, me whakaoti te whārite y=\frac{-19±11}{20} ina he tāpiri te ±. Tāpiri -19 ki te 11.

$$y=-\frac{8}{20}$$

Whakahekea te hautanga \frac{-8}{20} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

$$y=-\frac{2}{5}$$

Nā, me whakaoti te whārite y=\frac{-19±11}{20} ina he tango te ±. Tango 11 mai i -19.

$$y=-\frac{30}{20}$$

Whakahekea te hautanga \frac{-30}{20} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 10.

$$y=-\frac{3}{2}$$

Tauwehea te kīanga taketake mā te whakamahi i te ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Me whakakapi te -\frac{2}{5} mō te x_{1} me te -\frac{3}{2} mō te x_{2}.

$$10y^{2}+19y+6=10\left(y-\left(-\frac{2}{5}\right)\right)\left(y-\left(-\frac{3}{2}\right)\right)$$

Whakamāmātia ngā kīanga katoa o te āhua p-\left(-q\right) ki te p+q.

$$10y^{2}+19y+6=10\left(y+\frac{2}{5}\right)\left(y+\frac{3}{2}\right)$$

Tāpiri \frac{2}{5} ki te y mā te kimi i te tauraro pātahi me te tāpiri i ngā taurunga. Ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$10y^{2}+19y+6=10\times \frac{5y+2}{5}\left(y+\frac{3}{2}\right)$$

Tāpiri \frac{3}{2} ki te y mā te kimi i te tauraro pātahi me te tāpiri i ngā taurunga. Ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$10y^{2}+19y+6=10\times \frac{5y+2}{5}\times \frac{2y+3}{2}$$

Whakareatia \frac{5y+2}{5} ki te \frac{2y+3}{2} mā te whakarea taurunga ki te taurunga me te tauraro ki te tauraro, ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$10y^{2}+19y+6=10\times \frac{\left(5y+2\right)\left(2y+3\right)}{5\times 2}$$

Whakareatia 5 ki te 2.

$$10y^{2}+19y+6=10\times \frac{\left(5y+2\right)\left(2y+3\right)}{10}$$

Whakakorea atu te tauwehe pūnoa nui rawa 10 i roto i te 10 me te 10.

$$10y^{2}+19y+6=\left(5y+2\right)\left(2y+3\right)$$

Aromātai

$$\left(2y+3\right)\left(5y+2\right)$$